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Fed [463]
3 years ago
13

A 5.0 g sample of calcium nitrate (Ca(NO3)2, M = 164) contaminated with silica (SiO2, M = 60.1) is found to contain 1.0 g calciu

m. What is the mass percent purity of calcium nitrate in the sample?
Chemistry
1 answer:
zhenek [66]3 years ago
4 0

Answer:

= 82%

Explanation:

Percentage purity is calculated by the formula;

% purity = (mass of pure chemical/total mass of sample) × 100

In this case;

1 mole of Ca(NO3)2 = 164 g

but; 164 g of Ca(NO3)2 = 40 g Ca

Therefore; mass of Ca(NO3)2 = 164 /40

                                                  = 4.1 g

Thus;

% purity of Ca(NO3)2 = (Mass of Ca(NO3)2/ mass of the sample)× 100

                                    = (4.1 g/ 5 g) × 100

                                    = 82%

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Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in dinitro
vitfil [10]

Answer : The partial pressure of O_2 is, 222.93 torr

Explanation :  

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k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{168.6min}

k=4.11\times 10^{-3}min^{-1}

Now we have to calculate the partial pressure of O_2

The balanced chemical reaction is:

                           2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)

Initial pressure   760                0             0

At eqm.             (760-2x)            4x            x

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{P_o}{P_t}

where,

k = rate constant

t = time passed by the sample  = 215 min

a = initial pressure of N_2O_5 = 760 torr

a - x = pressure of N_2O_5 at equilibrium = (760-2x) torr

Now put all the given values in above equation, we get:

215=\frac{2.303}{4.11\times 10^{-3}}\log\frac{760}{760-2x}

x=222.93torr

The partial pressure of O_2 = x = 222.93 torr

7 0
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NaCl and AgNO3 react in a double replacement reaction. Which is one of the products of this reaction?
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Read 2 more answers
A 0.271g sample of an unknown vapor occupies 294ml at 140C and 874mmHg. The emperical formula of the compound is CH2. How many m
Phoenix [80]
Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
n = 116,524 x 294 x 10⁻⁶ / 8.314 x (140 + 273)
n = 9.98 x 10⁻³ mol

moles = mass / Mr
Mr = 0.271/9.98 x 10⁻³
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Formula of substance:
C₂H₄

Combustion equation:
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1 mole produces 2 moles of CO₂, so 3 moles will produce 6 moles CO₂
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